Answer
$4.1$ seconds
Work Step by Step
The ball hits the ground when $s(t)=0$.
We have to solve the quadratic equation:
$$-16t^2+64t+5=0.$$
The equation is in the standard form.
To solve the equation $ax^2+bx+c=0$ we will use the quadratic formula:
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Identify $a$, $b$, $c$:
$$\begin{align*}
a&=-16\\
b&=64\\
c&=5.
\end{align*}$$
We solve the given equation by substituting the values of $a$, $b$, $c$ in the quadratic formula:
$$\begin{align*}
t&=\dfrac{-64\pm\sqrt{64^2-4(-16)(5)}}{2(-16))}\\
&\approx \dfrac{-64\pm 66.45}{-32}\\
t_1&=\dfrac{-64-66.45}{-32}\approx 4.08\\
t_2&=\dfrac{-64+66.45}{-32}\approx -0.08.\\
\end{align*}$$
Because $t>0$ the only solution is $t\approx 4.1$ seconds.
